# The integrals converge. Evaluate the integrals without using tables \int_{0}^{1}\frac{4rdr}{\sqrt{1-r^{4}}}

The integrals converge. Evaluate the integrals without using tables
${\int }_{0}^{1}\frac{4rdr}{\sqrt{1-{r}^{4}}}$
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Step 1: Given that
The integrals converge. Evaluate the integrals without using tables
${\int }_{0}^{1}\frac{4rdr}{\sqrt{1-{r}^{4}}}$
Step 2: Solve
We have,
${\int }_{0}^{1}\frac{4rdr}{\sqrt{1-{r}^{4}}}dr={\int }_{0}^{1}\frac{4r}{\sqrt{1-{\left({r}^{2}\right)}^{2}}}dr$
Substitute,
${r}^{2}=t$
2r.dr=dt
$dr=\frac{dt}{2r}$
Plugging all the values into the integral we obtain,
${\int }_{0}^{1}\frac{4r}{\sqrt{1-{\left({r}^{2}\right)}^{2}}}dr{\int }_{0}^{1}\frac{4r}{\sqrt{1-{t}^{2}}}×\frac{dt}{2r}$
$=2{\int }_{0}^{1}\frac{dt}{\sqrt{1-{t}^{2}}}$
$=2{\left[{\mathrm{sin}}^{-1}\left(t\right)\right]}_{0}^{1}$
$=2\left({\mathrm{sin}}^{-1}\left(1\right)-0\right)$
$=2{\mathrm{sin}}^{-1}\mathrm{sin}\left(\frac{\pi }{2}\right)$
$=2×\frac{\pi }{2}$
$=\pi$
Jeffrey Jordon